How do I find the Laplace transformation of f(t) with varying intervals?

Karmel
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Homework Statement


fing the laplace transformation of f(t) where
-1 if 0<t<2
f(t)= e^3t if 2<t<4
2t if 4<t

Homework Equations





The Attempt at a Solution



I can get to the part where f(t) = -1 + (e^3t +1)u(t-2) + (2t-e^3t)u(t-4)
then I get lost ...Any help is appreciated. Thank you all in advance
 
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Hi Karmel,

First note that the -1 you have there should be -u(t); also, a minor point, but I'll assume u(t) is the Heaviside function restricted to t>0 (since f is only defined on t>0).

Your task now is to work with the formula g(t-a)u(t-a)\mapsto e^{-as}F(s) to take the Laplace transform of f(t). For example, we may rewrite the second term as

(e^{3t} + 1)u(t-2) = (e^6e^{3(t-2)} + 1)u(t-2),

and so by the above formula this transforms to e^{-2s}\left(e^6\frac{1}{s-3} + \frac{1}{s}\right).

Your turn. Show us your work if you get stuck.
 
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Okay so by using the examples from the book and the above example for the last part of the above problem I get and if I am right it will be luck!

(2t-e^3t)u(t-4) = (2t-(e^12)(e^3(t-4))u(t-4) which transforms into e^-4s(-e^12((1/s-3)+(2/s^2))

so assuming that this part is right. Do I leave the answer as is or do I need to multiply throughout and see if I can add anything and simplyfy it..

karmel
 
Almost, but you haven't completely made the factor multiplied by u(t-4) look like a function g(t-4) like in formula I gave above. We need to write it as

(2t - e^{3t})u(t - 4) = (2(t - 4) + 8 - e^{12}e^{3(t - 4)})u(t - 4).

so assuming that this part is right. Do I leave the answer as is or do I need to multiply throughout and see if I can add anything and simplyfy it..
Once you have transformed each term of f(t), then you have found the Laplace transform of f(t), as required.
 
Thank you so much for the help and assuming I haven't made any errors I do believe I have finally arrivied at the right answer. Once again thank you karmel
 
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